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Math Problem of the Day

(Updated 1/29/2016 2:22:04 AM)

Welcome to our daily solved problem section. Here you'll find daily solved problems. The topics are varied: Calculus, Algebra, Differential Equations, etc.

Do you have a suggestion, a brief problem you wanna get solved? Well, all you have to do is e-mail us. And we'll try to solve it as soon as possible.

Problem 1. Compute the derivative of the function

$\displaystyle f(x) = \frac{\sin x+x}{x - \cos x} $

Solution. We compute the derivative using the Quotient Rule. Notice that

$\displaystyle \frac{d}{dx} f(x) = \frac{d}{dx} \left( \frac{\sin x+x}{x - \cos
x}\right) = \frac{(\sin x +x )' (x-\cos x) - (\sin x+x)(x-\cos
x)'}{(x-\cos x)^2} $

$\displaystyle = \frac{(\cos x+1) (x-\cos x) - (\sin x+x)(1+\sin x)}{(x-\cos x)^2} $

$\displaystyle = \frac{ x\cos x+x - \cos^2 x -\cos x -
(\sin^2 x+x\sin x+\sin x+x) }{ (x-\cos x)^2 } $

$\displaystyle = \frac{ x\cos x+x - \cos^2 x -\cos x -
\sin^2 x-x\sin x-\sin x-x }{ (x-\cos x)^2 } $

$\displaystyle = \frac{ -\cos^2 x-\sin^2 x+ x\cos x -x\sin x-\sin x-\cos x }{ (x-\cos x)^2 } $

$\displaystyle = \frac{ -1+ x(\cos x -\sin x)-(\sin x+\cos x) }{ (x-\cos x)^2 } $

We could try to keep simplifying, but it'snot clear if we could make it more compact, so we just leave it like that.$ \Box$

Problem 2. Differentiate the function

$\displaystyle f(x) = x^4 \sec x \tan x $

Solution. We are going to use use the Product Rule, which yields

$\displaystyle \frac{d}{dx} \left( x^4 \sec x \tan x \right) = \frac{d}{dx} \lef...
...dx} \left(\sec x \right) \tan x + x^4 \sec x
\frac{d}{dx} \left(\tan x \right) $

Here we are using a generalized version of the Product Rule for three functions:

$\displaystyle (fgh)'=f'gh+fg'h+fgh' $

Pretty easy to remember, uh? Let's now come back to our example. Using the previous equality yields

$\displaystyle \frac{d}{dx} \left( x^4 \sec x \tan x \right) = 4x^3 \sec x \tan ...
...} \right) \tan x + x^4 \sec
x \frac{d}{dx} \left(\frac{\sin x}{\cos x} \right) $

$\displaystyle = 4x^3 \sec x \tan x + x^4 \frac{\sin x}{\cos^2 x} \tan x + x^4
\sec x \frac{1}{\cos^ 2x} $

$\displaystyle = 4x^3 \sec x \tan x + x^4 \sec x \tan^2 x + x^4
\sec^3 x $

$\displaystyle = x^3\sec x \left( 4\tan x + x\tan^2 x+ x\sec^2 x \right)$

$ \Box$

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